Annual Equivalent Worth Criterion Lecture No.19 Chapter Contemporary Engineering Economics Copyright © 2016 th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Chapter Opening Story: Robots May Revolutionize China’s Electronics Manufacturing o o o Cost of a robot: $10,000 Planning horizon: 20 years Cost of operating and owning the robot per year? Issue: Replacing people with robots would reduce the operating cost at the expense of increasing capital cost th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Annual Worth Analysis Principle: Measure an investment’s worth on an annual basis Benefits: By knowing the annual equivalent worth, we can: oSeek consistency of report format oDetermine the unit cost (or unit profit) oFacilitate the unequal project life • Annual Equivalent Conversion comparison th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Fundamental Decision Rules For Single Project: For Mutually Exclusive Alternatives: If AE(i) > 0, accept the investment Service projects: Select the alternative with If AE(i) = 0, remain indifferent to the investment the minimum annual equivalent cost (AEC) If AE(i) < 0, reject the investment Revenue projects: Select the alternative with the maximum AE(i) th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Example 6.1: Economics of Installing a Feed-water Heater • Install a 150MW unit • Initial cost = $1,650,000 • Service life = 25 years • Salvage value = • Expected improvement in fuel efficiency = 1% • Fuel cost = $0.05kWh • Load factor = 85% • Determine the annual worth for installing the unit at i = 12% • If the fuel cost increases at the annual rate of 4%, what is AE(12%)? th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Solution: Calculation of Annual Fuel Savings • Required input power before adding the second unit 150,000kW = 272,727kW 0.55 • Required input power after adding the second unit 150,000kW = 267,857kW 0.56 : • Reduction in energy consumption 272,727kW − 267,857kW = 4,870 kW • Annual operating hours Operating hours = (365)(24)(0.85) Annual Fuel Savings Afuel savings = (reduction in fuel requirement) × (fuel cost) ×(operating hours per year) =(4,870kW) × ($0.05/kWh) ×( 7,446 hours/year ) =$1,813,101/year =7,446 hours/year th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Solution: Annual Worth Calculations (a) with constant fuel price Cash Flow Diagrams PW(12%) = −$1,650,000 + $1,813,101(P / A,12%,25) = $12,570,403 AE(12%) = $12,573,321(A / P ,12%,25) = $1,602,726 (b) with escalating fuel price A1 =$1,813,101 PW (12%) = −$1,650,000 + $1,813,101(P / A1 ,4%,12%,25) = $17,459,783 AE (12%) = $17,459,783(A / P ,12%,25) = $2,226,122 th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Repeating Cash Flow Cycles • First Cycle PW(12%) = −$1,000,000 +[($800,000 − $100,000(A G ,12%,4)](P A ,12%,4) = −$1,000,000 + $2,017,150 = $1,017,150 • Repeating Cycles AE ( 12% ) = $1,017,150 ( A P , 12%, ) = $334,880 th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Example 6.3: Comparing Alternatives th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Solution • Required assumptions • • • The service life of the selected alternative is required on a continuous basis Each alternative will be replaced by an identical asset that has the same costs and performance Model A PW ( 15% ) = −$22,601 AEC ( 15% ) = $22,601 ( A P , 15%, ) • = $9,899 Model B PW ( 15% ) = −$25,562 AEC ( 15% ) = $25,562 ( A P , 15%, ) = $8,954 th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Annual Equivalent Cost (AEC) method is called the annual equivalent cost (AEC) Revenues must cover two kinds of costs: operating costs and capital costs Annual Equivalent Costs When only costs are involved, the AE Capital costs + Operating costs Annual equivalent cost = Capital cost + Operating costs th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Capital (Ownership) Cost • Def: Owning equipment associated with two transactions—(1) its initial cost (I), and (2) its salvage value (S) • Capital costs: Taking these items into consideration, we calculate the capital costs as: CR(i) = I(A / P , i , N) − S(A / F , i , N) = (I − S)(A / P , i , N) + iS th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Cost of Owning a Vehicle SEGMENT BEST MODELS ASKING PRICE PRICE AFTER YEARS Compact car Mini Cooper $19,800 $12,078 Midsize car Volkswagen Passat $28,872 $15,013 Sports car Porsche 911 $87,500 $48,125 Compact Luxury car BMW Series $39,257 $20,806 Luxury car Mercedes CLK $51,275 $30,765 Minivan Honda Odyssey $26,876 $15,051 Subcompact SUV Honda CR-V $20,540 $10,681 Compact SUV Acura MDX $37,500 $21,375 Full size SUV Toyota Sequoia $37,842 $18,921 Compact truck Toyota Tacoma $21,200 $10,812 Full size truck Toyota Tundra $25,653 $13,083 th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Example: Capital Cost (Mini-Cooper) Capital Recovery Cost Given: o I = $19,800 o N = years o S = $12,078 o i = 6% Find: CR(6%) CR(i) = (I -S) (A/P , i , N) + iS CR(6%) = ($19,800 - $12,078) (A/P , 6%, 3) + (0.06)$12,078 = $3,614 th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Example 6.4: Required Annual Revenue Cost of Owning and Operating Given: o o o o I = $20,000 S = $4,000 N = years i = 10% Find: See if an annual revenue of $5,000 is large enough to cover both the capital and operating costs th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Solution Need additional revenue in the amount of $120.76 to justify the investment th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved ... Pearson Education, Inc All Rights Reserved Annual Worth Analysis Principle: Measure an investment’s worth on an annual basis Benefits: By knowing the annual equivalent worth, we can: oSeek consistency... All Rights Reserved Annual Equivalent Cost (AEC) method is called the annual equivalent cost (AEC) Revenues must cover two kinds of costs: operating costs and capital costs Annual Equivalent Costs... Reduction in energy consumption 272,727kW − 267,857kW = 4,870 kW • Annual operating hours Operating hours = (365)(24)(0.85) Annual Fuel Savings Afuel savings = (reduction in fuel requirement)